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A 2.00-l reaction vessel, initially at 298 k, contains chlorine gas at a partial pressure of 337 mmhg and fluorine gas at a partial pressure of 729 mmhg. identify the limiting reactant and determine the theoretical yield of clf3 in grams.

Respuesta :

It is 189 k of clf3 ok ok ok ok ok ok ok ok ok ok ok

Answer: The theoretical yield of [tex]ClF_3[/tex] is 2.24 g

Explanation:

The equation given by ideal gas follows:

[tex]PV=nRT[/tex]          .......(1)

where, P = pressure of the gas

V = Volume of the gas

T = Temperature of the gas

R = Gas constant = [tex]62.3637\text{ L mmHg }mol^{-1}K^{-1}[/tex]

n = number of moles of gas

  • For fluorine gas:

We are given:

[tex]P=337mmHg\\V=2.00L\\T=298K\\n=?[/tex]

Putting values in equation 1, we get:

[tex]337mmHg\times 2.00L=n\times 62.3637\text{ L mmHg }mol^{-1}K^{-1}\times 298K\\\\n=\frac{337\times 2.00}{62.3637\times 298}=0.0363mol[/tex]

  • For chlorine gas:

We are given:

[tex]P=729mmHg\\V=2.00L\\T=298K\\n=?[/tex]

Putting values in equation 1, we get:

[tex]729mmHg\times 2.00L=n\times 62.3637\text{ L mmHg }mol^{-1}K^{-1}\times 298K\\\\n=\frac{729\times 2.00}{62.3637\times 298}=0.0784mol[/tex]

The chemical equation for the reaction of chlorine gas and fluorine gas follows:

[tex]Cl_2+3F_2\rightarrow 2ClF_3[/tex]

By Stoichiometry of the reaction:

3 moles of fluorine gas reacts with 1 mole of chlorine gas.

So, 0.0363 moles of fluorine gas will react with = [tex]\frac{1}{3}\times 0.0363=0.0121mol[/tex] of chlorine gas

As, given amount of chlorine gas is more than the required amount. So, it is considered as an excess reagent.

Thus, fluorine gas is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

3 moles of fluorine gas produces 2 moles of [tex]ClF_3[/tex]

So, 0.0363 moles of fluorine gas will produce = [tex]\frac{2}{3}\times 0.0363=0.0242mol[/tex] of [tex]ClF_3[/tex]

  • To calculate the theoretical yield of [tex]ClF_3[/tex] we use the equation:

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]

Molar mass of [tex]ClF_3[/tex] = 92.45 g/mol

Moles of [tex]ClF_3[/tex] = 0.0242 moles

Putting values in above equation, we get:

[tex]0.0242mol=\frac{\text{Mass of }ClF_3}{92.45g/mol}\\\\\text{Mass of }ClF_3=(0.0242mol\times 92.45g/mol)=2.24g[/tex]

Hence, the theoretical yield of [tex]ClF_3[/tex] is 2.24 g

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